176 Math 216: Foundations of

8.4 Images of morphisms: Chevalley’s theorem and elimination theory

In this section, we will answer a question that you may have wondered about long before hearing the phrase “algebraic geometry”. If you have a number of equations in a number of variables with indeterminate coefficients, and you wonder if they have a solution, you would reasonably ask what condi- tions there are on the coefficients for a common solution to exist. Given the al- gebraic nature of the problem, you might hope that the answer should be purely algebraic in nature — it shouldn’t be “random”, or involve bizarre functions like exponentials or cosines. This is indeed the case, and was first understood via the nineteenth century theory of “elimination of variables”. It can be profitably inter- preted as a question about images of maps of varieties or schemes, in which guise it is answered by Chevalley’s theorem.

8.4.1. Images of morphisms, and Chevalley’s theorem. If f : X Y is a morphism of schemes, the notion of the image of f as sets is clear: we just take the points in Y that are the image of points in X. We know that the image→ can be open (open immersions), and we will soon see that it can be closed (closed immersions), and hence locally closed (locally closed immersions). But it can be weirder still: consider the morphism A2 A2 given by (x, y) k k ! (x, xy). The image is the plane, with the x-axis removed, but the origin put back in. This isn’t so horrible. We make a definition to capture→ this phenomenon.→ A constructible subset of a Noetherian topological space is a subset which belongs to the smallest family of subsets such that (i) every open set is in the family, (ii) a finite intersection of family members is in the family, and (iii) the complement of a family member is also in the family. For example the image of (x, y) (x, xy) is ! constructible. → 8.4.A. EXERCISE: CONSTRUCTIBLESUBSETSINTERMSOFLOCALLYCLOSEDSUB- SETS. Suppose X is a topological space. Recall that a subset of X is locally closed if it is the intersection of an open subset and a closed subset. (Equivalently, it is an open subset of a closed subset, or a closed subset of an open subset. We will later have trouble extending this to open and closed and locally closed subschemes, see Exercise 9.1.K.) Show that a subset of X is constructible if and only if it is the finite disjoint union of locally closed subsets. As a consequence, if X Y is a continuous map of topological spaces, then the preimage of a constructible set is a constructible set. → One nice property of constructible subsets of schemes is that there is a short criterion for openness: a constructible subset is open if it is “closed under gener- ization” (see Exercise 24.2.N). The image of a morphism of schemes can be stranger than constructible. In- deed if S is any subset of a scheme Y, it can be the image of a morphism: let X be the disjoint union of spectra of the residue fields of all the points of S, and let f : X Y be the natural map. This is quite pathological, but in any reasonable situation, the image is essentially no worse than arose in the previous example of (x, y) (x, xy). This is made precise by Chevalley’s theorem. →! → November 22, 2010 draft 177

8.4.2. Chevalley’s theorem. — If π : X Y is a finite type morphism of Noetherian schemes, the image of any constructible set is constructible. In particular, the image of π is constructible. → To prove Chevalley’s theorem, we prove the following classical result.

8.4.3. Theorem (Fundamental Theorem of Elimination Theory). — The morphism n π : PA Spec A is closed (sends closed sets to closed sets). Theorem 8.4.3 is simpler than Chevalley’s theorem; and there are only equali- ties, no→ inequalities. A great deal of classical algebra and geometry is contained in it as special cases. Here are some examples. 2 First, let A = k[a, b, c, . . . , i], and consider the closed subscheme of PA (taken with coordinates x, y, z) corresponding to ax + by + cz = 0, dx + ey + fz = 0, gx + hy + iz = 0. Then we are looking for the locus in Spec A where these equations have a non-trivial solution. This indeed corresponds to a Zariski-closed set — where

a b c det d e f = 0. g h i   Thus the idea of the is embedded in elimination theory. As a second example, let A = k[a0,a1, . . . , am,b0,b1, . . . , bn]. Now consider 1 the closed subscheme of PA (taken with coordinates x and y) corresponding to a xm + a xm−1y + + a ym = 0 and b xn + b xm−1y + + b yn = 0. 0 1 ··· m 0 1 ··· n Then there is a polynomial on the coefficients a0,..., bn (an element of A) which vanishes if and only if these two have a common non-zero root — this polynomial is called the . More generally, this question boils down to the following question. Given a number of homogeneous equations in n + 1 variables with indeterminate coeffi- cients, Proposition 8.4.3 implies that one can write down equations in the coeffi- cients that will precisely determine when the equations have a nontrivial solution.

n Proof of the Fundamental Theorem of Elimination Theory 8.4.3. Suppose Z ! PA is a closed subset. We wish to show that π(Z) is closed. (See Figure 8.6.) Suppose y/π(Z) is a closed point of Spec A. We will check that there is a ∈ → distinguished open neighborhood D(f) of y in Spec A such that D(f) doesn’t meet π(Z). (If we could show this for all points of π(Z), we would be done. But I prefer to concentrate on closed points first for simplicity.) Suppose y corresponds to the maximal ideal m of A. We seek f A −m such that π∗f vanishes on Z. ∈ n −1 Let U0,..., Un be the usual affine open cover of PA. The closed subsets π y and Z do not intersect. On the affine open set Ui, we have two closed subsets Z U and π−1y U that do not intersect, which means that the ideals corre- ∩ i ∩ i sponding to the two closed sets generate the unit ideal, so in the ring of functions A[x0/i,x1/i, . . . , xn/i]/(xi/i − 1) on Ui, we can write

1 = ai + mijgij " 178 Math 216: Foundations of Algebraic Geometry

−1 n Z π y PA

π

D(f) Spec A y

FIGURE 8.6.

where mij m, and ai vanishes on Z. Note that ai,gij A[x0/i, . . . , xn/i]/(xi/i − ∈ n ∈ 1), so by multiplying by a sufficiently high power xi of xi, we have an equality N xi = ai" + mijgij" in S = A[x0, . . . , xn]. We may take N large" enough so that it works for all i. Thus • for N " sufficiently large, we can write any monomial in x1,..., xn of degree N" as something vanishing on Z plus a linear combination of elements of m times other polynomials. Hence

SN ! = I(Z)N ! +mSN ! where I(Z) is the graded ideal of functions vanishing on Z. By Nakayama’s • lemma (version 1, Lemma 8.2.8), taking M = SN ! /I(Z)N ! , we see that there ex- ists f A −m such that ∈ fS I(Z) . N ! ⊂ N ! Thus we have found our desired f. We now tackle Theorem 8.4.3 in general, by simply extending the above argu- ment so that y need not be a closed point. Suppose y = [p] not in the image of Z. Applying the above argument in Spec A , we find S A = I(Z) A + p N ! ⊗ p N ! ⊗ p mS A , from which g(S /I(Z) ) A = 0 for some g A −pA , from N ! ⊗ p N ! N ! ⊗ p ∈ p p which (S /I(Z) ) A = 0. As S is a finitely generated A-module, there is N ! N ! ⊗ p N ! some f A −p with fS I(Z) (if the module-generators of S , and f ,..., f ∈ N ⊂ N ! 1 a are annihilate the generators respectively, then take f = fi), so once again we have found D(f) containing p, with (the pullback of) f vanishing on Z. ! # Notice that projectivity was crucial to the proof: we used graded rings in an essential way. Proof of Chevalley’s Theorem 8.4.2. We begin with a series of reductions.

8.4.B. EXERCISE. (a) Reduce to the case where Y is affine, say Y = Spec B. (b) Reduce further to the case where X is affine. November 22, 2010 draft 179

n (c) Reduce further to the case where X = AB = Spec B[t1, . . . , tn]. 1 (d) By induction on n, reduce further to the case where X = AB = Spec B[t]. (e) Reduce to showing that for any Noetherian ring B, and any locally closed sub- set Z A1 , the image of Z under the projection π : A1 Spec B is constructible. ⊂ B B (f) Reduce to showing that for any Noetherian ring B, and any irreducible closed sub- set Z A1 , the image of Z under the projection π : A1 Spec B is constructible. ⊂ B B→

1 → Now Z is cut out (in A ) by (finitely many) polynomials f1, . . . , fm B[t]. Let B ∈ F be the subset of Spec B consisting of points of Spec B where Z contains the entire fiber above such points.

8.4.C. EXERCISE. If a ,..., a B are the coefficients of the polynomials 1 M ∈ f1, . . . , fm, show that F = V(a1, . . . , aM), and hence that F is closed.

8.4.D. EXERCISE. By covering Spec B \ F with finitely many affines (Spec B \ F is quasicompact by Exercise 4.6.M), show that it suffices to prove the statement given in Exercise 8.4.B(f) in the case where Z contains no fibers of π. 1 1 We now consider AB as a standard open subset of PB, with complement H 1 (“the hyperplane at infinity”). We still denote the structure morphism PB 1 Spec B by π. Let Z be the closure of Z in PB, and let Z = Z \ Z. By the Fun- damental Theorem of Elimination Theory 8.4.3, π(Z ) and π(Z) are both closed→ in Spec B. ∞ ∞ 8.4.E. EXERCISE. Show that π(Z) is irreducible.

8.4.F. SUBTLE EXERCISE. Show that π(Z ) is strictly contained in π(Z). Hint: show that the fiber above the generic point of π(Z) cannot meet Z . Using Noetherian induction on B, we may∞ assume the desired result over any proper closed subset of Spec B, so applying this to π(Z ), we see that∞ π(Z) π(Z ) ∩ is constructible. But also π(Z) \ π(Z )=π(Z) \ π(Z ) is also constructible (it is ∞ ∞ the difference of two closed sets), so we are done. ! ∞ ∞

8.4.G. "" EXERCISE (CHEVALLEY’S THEOREM FOR LOCALLY FINITELY PRESENTED MORPHISMS). If you are macho and are embarrassed by Noetherian rings, the following extension of Chevalley’s theorem will give you a sense of one of the standard ways of removing Noetherian hypotheses. (a) Suppose that A is a finitely presented B-algebra, so A = B[x1, . . . , xn]/(f1, . . . , fr). Show that the image of Spec A Spec B is a constructible subset of Spec B. Hint: describe Spec A Spec B as the base change of → Spec Z[x→1, . . . , xn,a1, . . . , aN]/(g1, . . . , gn) Spec Z[a1, . . . , aN], where the images of ai in Spec B are the coefficients→ of the fj (there is one ai for each coefficient of each f ), and g f . j i ! i (b) Show that if π : X Y is a quasicompact locally finitely presented morphism, and Y is quasicompact, then π(X)→is constructible. (For hardened experts only: [EGA,0III.9.1] gives a→ definition of constructability, and local constructability, in more generality. The general form of Chevalley’s constructibility theorem [EGA, 180 Math 216: Foundations of Algebraic Geometry

IV1.1.8.4] is that the image of a locally constructible set, under a finitely presented map, is also locally constructible.)

8.4.4. " Elimination of quantifiers. A basic sort of question that arises in any number of contexts is when a system of equations has a solution. Suppose for example you have some polynomials in variables x1,..., xn over an algebraically closed field k, some of which you to set to be zero, and some of which you set to be nonzero. (This question is of fundamental interest even before you know any scheme theory!) Then there is an algebraic condition on the coefficients which will n tell you if this is the case. Define the Zariski topology on k in the obvious way: closed subsets are cut out by equations.

8.4.H. EXERCISE (ELIMINATION OF QUANTIFIERS, OVERANALGEBRAICALLYCLOSED FIELD). Fix an algebraically closed field k. Suppose

f1, . . . , fp,g1, . . . , gq k[A1,...,Am,X1,...Xn] ∈ m are given. Show that there is a Zariski-constructible subset Y of k such that (8.4.4.1) f (a , . . . , a ,X ,...,X )= = f (a , . . . , a ,X ,...,X )=0 1 1 m 1 n ··· p 1 m 1 n and

(8.4.4.2) g1(a1, . . . , am,X1,...,Xn) = 0 gp(a1, . . . , am,X1,...,Xn) = 0 & ··· n & has a solution (X ,...,X ) = (x , . . . , x ) k if and only if (a , . . . , a ) Y. 1 n 1 n ∈ 1 m ∈ Hints: if Z is a finite type scheme over k, and the closed points are denoted Zcl (“cl” is for either “closed” or “classical”), then the inclusion of topological spaces Zcl ! Z, the Zariski topology on Z induces the Zariski topology on Zcl. Note that p we can identify ( p)cl with k by the Nullstellensatz (Exercise 6.3.E). If X is the Ak locally→ closed subset of Am+n cut out by the equalities and inequalities (8.4.4.1) and (8.4.4.2), we have the diagram   l. cl. imm. Xcl / X / Am+n ww cl π ww π ww ww m   {w k / Am where Y = im πcl. By Chevalley’s theorem 8.4.2, im π is constructible, and hence m m so is (im π) k . It remains to show that (im π) k = Y (= im πcl). You might ∩ ∩ use the Nullstellensatz. This is called “elimination of quantifiers” because it gets rid of the quantifier “there exists a solution”. The analogous statement for real numbers, where inequal- ities are also allowed, is a special case of Tarski’s celebrated theorem of elimination of quantifiers for real closed fields.